Question

In the given figure, ABCD is a trapezium. Find the sum of areas of $△AOD$ and $△COB$.

• A. $10c{m}^2$
• B. $25c{m}^2$
• C. $20c{m}^2$
• D. $45c{m}^2$

Answer 1

The correct option is C $20c{m}^2$



In $△AOD$ and $△COB$,
$\angle A=\angle C$ ( Alternate interior angles)
$\angle B=\angle D$ ( Alternate interior angles)
$\angle AOB=\angle COD$ (vertically opposite angles)
By AAA criteria,
$△AOB\sim △COD$

Let’s draw MN passing through O and perpendicular to parallel sides of trapezium.

$\frac{Areaof△AOB}{Areaof△COD}=\frac{A{B}^2}{C{D}^2}\Rightarrow \frac{\frac{1}{2}\times AB\times ON}{\frac{1}{2}\times CD\times OM}=\frac{A{B}^2}{C{D}^2}$

$\frac{ON}{OM}=\frac{AB}{CD}\Rightarrow \frac{ON}{OM}=\frac{10}{5}=\frac{2}{1}$

$ON+OM=6cm$
$ON=4cm$ and $OM=2cm$

$Areaof△AOB+Areaof△COD=\frac{1}{2}\times AB\times ON+\frac{1}{2}\times CD\times OM$

$Areaof△AOB+Areaof△COD=\frac{1}{2}\times 10\times 4+\frac{1}{2}\times 5\times 2=25c{m}^2$

Area of trapezium = $\frac{1}{2}\times (AB+CD)\times MN=\frac{1}{2}\times (10+5)\times 6=45c{m}^2$

$Areaof△AOD+Areaof△COB=\text{Area of trapezium}-(Areaof△AOB+Areaof△COD)$

$Areaof△AOD+Areaof△COB=45c{m}^2-25c{m}^2=20c{m}^2$